Let $F$ be a subspace of a symplectic space $P$, we denote by $F^{\circ}$ a symplectic orthogonal of $F$.
Why is $(A+B)^{\circ} = A^{\circ}\cap B^{\circ}$, or where can I find the proof of this fact?
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Are $A$ and $B$ subspaces of $P$? – Michael Albanese Mar 09 '13 at 11:37
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Actually, in the textbook that I am reading it is written independently, but it is applied for subspaces of $P$. – Novak Djokovic Mar 09 '13 at 11:43
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1As you haven't specified what is meant by the symplectic orthogonal of $A$ where $A$ is not a subspace, I have taken $A$ and $B$ to be subspaces in my answer below. – Michael Albanese Mar 09 '13 at 11:47
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Is my answer acceptable? If not, let me know what it is you are yet to understand. – Michael Albanese Sep 27 '13 at 19:00
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yes, sure. thanks a lot :) – Novak Djokovic Sep 30 '13 at 02:59
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If you find it acceptable, please accept the answer so that the question is no longer left open. – Michael Albanese Sep 30 '13 at 03:31
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sorry, but i dont see that option, i can just vote if i find it useful, but i did that before.. – Novak Djokovic Sep 30 '13 at 08:25
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There is a tick under the up/down vote buttons. Have a look at How do I accept an answer?. You should also go through all the questions you've asked before; if there is an acceptable answer on any of them, make sure to accept it, but if there isn't, leave it open. – Michael Albanese Sep 30 '13 at 12:11
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If $v \in (A + B)^{\circ}$ then $\omega(v, a + b) = 0$ for all $a \in A$ and $b \in B$. Choosing $a = 0$ (noting that $0 \in A$ as it is a subspace), we have $\omega(v, b) = 0$ for all $b \in B$, so $v \in B^{\circ}$. Similarly, choosing $b = 0$ instead, we have $v \in A^{\circ}$. Therefore $v \in A^{\circ}\cap B^{\circ}$, so $(A + B)^{\circ} \subseteq A^{\circ}\cap B^{\circ}$.
The reverse direction is a little bit easier. I recommend you try it yourself, but if you're stuck, just put your mouse over the paragraph below.
If $v \in A^{\circ}\cap B^{\circ}$, then $\omega(v, a) = 0$ and $\omega(v, b) = 0$ for all $a \in A$ and for all $b \in B$. Adding the two equations and using the bilinearity of $\omega$, we have $\omega(v, a + b) = 0$ for all $a \in A$ and for all $b \in B$. Therefore $v \in (A+B)^{\circ}$, so $A^{\circ}\cap B^{\circ} \subseteq (A+B)^{\circ}$.
Hence $(A+B)^{\circ} = A^{\circ}\cap B^{\circ}$.
Michael Albanese
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